As soon as the oxide of general material M is heated at 25C, just a

minimal number of M is created.

MO2(s) <---> M(s)+O2(g)

delta G = 288.5 kJ/mol 1.) As soon as the response is paired into the transformation of graphite

to carbon-dioxide, it becomes natural.

What’s the chemical equation with this paired procedure? Show that

the response is within balance, consist of actual says, and

represent graphite as C(S) _________________________________

2.) what’s the thermodynamic balance continual the paired

response?

K = ????

(1) (A) MO2(s) <=> M(s) + O2(g), Delta G1 = 288.5

kJ/mol

(B) C(s) + O2(g) <=> CO2(g), Delta G2 = -394.4 kJ/mol

Include equation (A) + equation (B):

MO2(s) + C(s) + O2(g) <=> M(s) + O2(g) + CO2(g)

Cancel typical terms for paired response:

MO2(s) + C(s)

<=> M(s) + CO2(g)

Delta Get = Delta G1 + Delta G2

= 288.5 + (-394.4)

= -105.9 kJ/mol = -105900 J/mol

(2) heat T = 25 deg C = 298.15 K

Molar fuel continual roentgen = 8.314 J/mol.K

Equilibrium continual K = exp(-Delta Go/RT)

= exp(105900/(8.314 x 298.15))

= 3.58 x

10^(18)

(Note: if you are using T = 298 K, after that K = 3.66 x

10^18)

O2M(s)+ C(s) —-> M(s) + CO2(g)

I’d include 288.9 to dGof for CO2 of -394.4 for -105.5 kJ/mol

the response you reveal. After That

(105,500/8.314*298) = lnK and I also have and endless choice for K. Somethng

like 3.1E18